\(\int \frac {\tan ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) [335]

Optimal result

Integrand size = 23, antiderivative size = 148 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {2 \left (a^2-b^2\right )^2}{a b^4 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (3 a^2-2 b^2\right ) \sqrt {a+b \sec (c+d x)}}{b^4 d}-\frac {2 a (a+b \sec (c+d x))^{3/2}}{b^4 d}+\frac {2 (a+b \sec (c+d x))^{5/2}}{5 b^4 d} \]

[Out]

-2*arctanh((a+b*sec(d*x+c))^(1/2)/a^(1/2))/a^(3/2)/d-2*a*(a+b*sec(d*x+c))^(3/2)/b^4/d+2/5*(a+b*sec(d*x+c))^(5/
2)/b^4/d+2*(a^2-b^2)^2/a/b^4/d/(a+b*sec(d*x+c))^(1/2)+2*(3*a^2-2*b^2)*(a+b*sec(d*x+c))^(1/2)/b^4/d

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3970, 912, 1275, 212} \[ \int \frac {\tan ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {2 \left (3 a^2-2 b^2\right ) \sqrt {a+b \sec (c+d x)}}{b^4 d}+\frac {2 \left (a^2-b^2\right )^2}{a b^4 d \sqrt {a+b \sec (c+d x)}}+\frac {2 (a+b \sec (c+d x))^{5/2}}{5 b^4 d}-\frac {2 a (a+b \sec (c+d x))^{3/2}}{b^4 d} \]

[In]

Int[Tan[c + d*x]^5/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

(-2*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]])/(a^(3/2)*d) + (2*(a^2 - b^2)^2)/(a*b^4*d*Sqrt[a + b*Sec[c + d*x
]]) + (2*(3*a^2 - 2*b^2)*Sqrt[a + b*Sec[c + d*x]])/(b^4*d) - (2*a*(a + b*Sec[c + d*x])^(3/2))/(b^4*d) + (2*(a
+ b*Sec[c + d*x])^(5/2))/(5*b^4*d)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 912

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{q = De
nominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + g*(x^q/e))^n*((c*d^2 + a*e^2)/e^2 - 2*c*
d*(x^q/e^2) + c*(x^(2*q)/e^2))^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*
g, 0] && NeQ[c*d^2 + a*e^2, 0] && IntegersQ[n, p] && FractionQ[m]

Rule 1275

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 3970

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[-(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[(b^2 - x^2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (b^2-x^2\right )^2}{x (a+x)^{3/2}} \, dx,x,b \sec (c+d x)\right )}{b^4 d} \\ & = \frac {2 \text {Subst}\left (\int \frac {\left (-a^2+b^2+2 a x^2-x^4\right )^2}{x^2 \left (-a+x^2\right )} \, dx,x,\sqrt {a+b \sec (c+d x)}\right )}{b^4 d} \\ & = \frac {2 \text {Subst}\left (\int \left (3 a^2 \left (1-\frac {2 b^2}{3 a^2}\right )-\frac {\left (a^2-b^2\right )^2}{a x^2}-3 a x^2+x^4-\frac {b^4}{a \left (a-x^2\right )}\right ) \, dx,x,\sqrt {a+b \sec (c+d x)}\right )}{b^4 d} \\ & = \frac {2 \left (a^2-b^2\right )^2}{a b^4 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (3 a^2-2 b^2\right ) \sqrt {a+b \sec (c+d x)}}{b^4 d}-\frac {2 a (a+b \sec (c+d x))^{3/2}}{b^4 d}+\frac {2 (a+b \sec (c+d x))^{5/2}}{5 b^4 d}-\frac {2 \text {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,\sqrt {a+b \sec (c+d x)}\right )}{a d} \\ & = -\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {2 \left (a^2-b^2\right )^2}{a b^4 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (3 a^2-2 b^2\right ) \sqrt {a+b \sec (c+d x)}}{b^4 d}-\frac {2 a (a+b \sec (c+d x))^{3/2}}{b^4 d}+\frac {2 (a+b \sec (c+d x))^{5/2}}{5 b^4 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.56 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.79 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 \left (5 b^4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+\frac {b \sec (c+d x)}{a}\right )+a \left (4 a \left (4 a^2-5 b^2\right )+2 b \left (4 a^2-5 b^2\right ) \sec (c+d x)-2 a b^2 \sec ^2(c+d x)+b^3 \sec ^3(c+d x)\right )\right )}{5 a b^4 d \sqrt {a+b \sec (c+d x)}} \]

[In]

Integrate[Tan[c + d*x]^5/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

(2*(5*b^4*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (b*Sec[c + d*x])/a] + a*(4*a*(4*a^2 - 5*b^2) + 2*b*(4*a^2 - 5*b^
2)*Sec[c + d*x] - 2*a*b^2*Sec[c + d*x]^2 + b^3*Sec[c + d*x]^3)))/(5*a*b^4*d*Sqrt[a + b*Sec[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1233\) vs. \(2(132)=264\).

Time = 15.60 (sec) , antiderivative size = 1234, normalized size of antiderivative = 8.34

method	result	size
default	\(\text {Expression too large to display}\)	\(1234\)

[In]

int(tan(d*x+c)^5/(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/5/d/a^2/b^4*(a+b*sec(d*x+c))^(1/2)/((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)/(b+a*cos(d*x+c))^2/
(cos(d*x+c)+1)*(5*cos(d*x+c)^3*ln(4*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^(1/2)+4*
a*cos(d*x+c)+4*a^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*b)*a^(5/2)*b^4+10*cos(d*x+c)^2*l
n(4*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^(1/2)+4*a*cos(d*x+c)+4*a^(1/2)*((b+a*cos
(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*b)*a^(3/2)*b^5-32*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)
^(1/2)*a^6*cos(d*x+c)^3+40*cos(d*x+c)^3*a^4*b^2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)-10*cos(d*
x+c)^3*a^2*b^4*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+5*cos(d*x+c)*ln(4*cos(d*x+c)*((b+a*cos(d*x
+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^(1/2)+4*a*cos(d*x+c)+4*a^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x
+c)+1)^2)^(1/2)+2*b)*a^(1/2)*b^6-32*cos(d*x+c)^2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^6-48*(
(b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^5*b*cos(d*x+c)^2+40*cos(d*x+c)^2*((b+a*cos(d*x+c))*cos(d
*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^4*b^2+60*cos(d*x+c)^2*a^3*b^3*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(
1/2)-10*cos(d*x+c)^2*a^2*b^4*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)-10*cos(d*x+c)^2*a*b^5*((b+a*
cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)-48*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1
/2)*a^5*b-12*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^4*b^2*cos(d*x+c)+60*cos(d*x+c)*((b+a*cos(d
*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^3*b^3+20*cos(d*x+c)*a^2*b^4*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+
c)+1)^2)^(1/2)-10*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a*b^5-12*((b+a*cos(d*x+c))*c
os(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^4*b^2+2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^3*b^3+20*((
b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^2*b^4+2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1
/2)*a^3*b^3*sec(d*x+c)-2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^2*b^4*sec(d*x+c)-2*((b+a*cos(d
*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^2*b^4*sec(d*x+c)^2)

Fricas [A] (verification not implemented)

none

Time = 0.50 (sec) , antiderivative size = 467, normalized size of antiderivative = 3.16 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\left [\frac {5 \, {\left (a b^{4} \cos \left (d x + c\right )^{3} + b^{5} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (-8 \, a^{2} \cos \left (d x + c\right )^{2} - 8 \, a b \cos \left (d x + c\right ) - b^{2} + 4 \, {\left (2 \, a \cos \left (d x + c\right )^{2} + b \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}\right ) - 4 \, {\left (2 \, a^{3} b^{2} \cos \left (d x + c\right ) - a^{2} b^{3} - {\left (16 \, a^{5} - 20 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (4 \, a^{4} b - 5 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{10 \, {\left (a^{3} b^{4} d \cos \left (d x + c\right )^{3} + a^{2} b^{5} d \cos \left (d x + c\right )^{2}\right )}}, \frac {5 \, {\left (a b^{4} \cos \left (d x + c\right )^{3} + b^{5} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + b}\right ) - 2 \, {\left (2 \, a^{3} b^{2} \cos \left (d x + c\right ) - a^{2} b^{3} - {\left (16 \, a^{5} - 20 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (4 \, a^{4} b - 5 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{5 \, {\left (a^{3} b^{4} d \cos \left (d x + c\right )^{3} + a^{2} b^{5} d \cos \left (d x + c\right )^{2}\right )}}\right ] \]

[In]

integrate(tan(d*x+c)^5/(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/10*(5*(a*b^4*cos(d*x + c)^3 + b^5*cos(d*x + c)^2)*sqrt(a)*log(-8*a^2*cos(d*x + c)^2 - 8*a*b*cos(d*x + c) -
b^2 + 4*(2*a*cos(d*x + c)^2 + b*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))) - 4*(2*a^3*b^2*
cos(d*x + c) - a^2*b^3 - (16*a^5 - 20*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^3 - 2*(4*a^4*b - 5*a^2*b^3)*cos(d*x + c)
^2)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)))/(a^3*b^4*d*cos(d*x + c)^3 + a^2*b^5*d*cos(d*x + c)^2), 1/5*(5*(a*
b^4*cos(d*x + c)^3 + b^5*cos(d*x + c)^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*co
s(d*x + c)/(2*a*cos(d*x + c) + b)) - 2*(2*a^3*b^2*cos(d*x + c) - a^2*b^3 - (16*a^5 - 20*a^3*b^2 + 5*a*b^4)*cos
(d*x + c)^3 - 2*(4*a^4*b - 5*a^2*b^3)*cos(d*x + c)^2)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)))/(a^3*b^4*d*cos(
d*x + c)^3 + a^2*b^5*d*cos(d*x + c)^2)]

Sympy [F]

\[ \int \frac {\tan ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\tan ^{5}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(tan(d*x+c)**5/(a+b*sec(d*x+c))**(3/2),x)

[Out]

Integral(tan(c + d*x)**5/(a + b*sec(c + d*x))**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.31 \[ \int \frac {\tan ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {\frac {5 \, \log \left (\frac {\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {10}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} a} + \frac {2 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}}}{b^{4}} - \frac {10 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a}{b^{4}} + \frac {30 \, \sqrt {a + \frac {b}{\cos \left (d x + c\right )}} a^{2}}{b^{4}} + \frac {10 \, a^{3}}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} b^{4}} - \frac {20 \, \sqrt {a + \frac {b}{\cos \left (d x + c\right )}}}{b^{2}} - \frac {20 \, a}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} b^{2}}}{5 \, d} \]

[In]

integrate(tan(d*x+c)^5/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

1/5*(5*log((sqrt(a + b/cos(d*x + c)) - sqrt(a))/(sqrt(a + b/cos(d*x + c)) + sqrt(a)))/a^(3/2) + 10/(sqrt(a + b
/cos(d*x + c))*a) + 2*(a + b/cos(d*x + c))^(5/2)/b^4 - 10*(a + b/cos(d*x + c))^(3/2)*a/b^4 + 30*sqrt(a + b/cos
(d*x + c))*a^2/b^4 + 10*a^3/(sqrt(a + b/cos(d*x + c))*b^4) - 20*sqrt(a + b/cos(d*x + c))/b^2 - 20*a/(sqrt(a +
b/cos(d*x + c))*b^2))/d

Giac [F]

\[ \int \frac {\tan ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {\tan \left (d x + c\right )^{5}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(tan(d*x+c)^5/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(tan(d*x + c)^5/(b*sec(d*x + c) + a)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^5}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

[In]

int(tan(c + d*x)^5/(a + b/cos(c + d*x))^(3/2),x)

[Out]

int(tan(c + d*x)^5/(a + b/cos(c + d*x))^(3/2), x)